3 questions about Algebraic Geometry and Zariski topology

Question

I've read this exercise on my book, but I don't know how to prove it... Can someone help me?

Let X $\subseteq \mathbb{A}^n$ be an algebraic set, and $f, g:X\rightarrow k$ two regular functions.

a) If we see $f$ as an application $X \rightarrow \mathbb{A}^1(k)$, prove that $f$ is continuous in the Zariski topology.

b) If $X$ is irreducible and an open $U \subseteq X$ exists such that $f_{|U} = g_{|U}$, prove that $f=g$.

c) Give an example of $X, f$ and $g$ such that $f \neq g$ but $f_{|U} = g_{|U}$ in an open $U \subseteq X$.

Thank you!

For b) I have been able to prove that:

Let E be the subset where $f = g$ so E is closed (Why? I know that $E = (f,g)^{-1}(\Delta_{k})$ but... why is $(\Delta_{k})$ closed?). So $U \subseteq E \Rightarrow \bar{U} \subseteq \bar{E}=E$ (beacuse E is closed) $\Rightarrow X \subseteq E$, because $\bar{U}=X \Rightarrow E = X$ and $f=g$ in X.

Answer 1

Here are a few facts to get you started.

1. $f$ is continous if and only if for every closed set $V$, the inverse image $f^{-1} (V)$ is closed. The closed sets of $\mathbb A^1$ are finite sets.
2. That $f=g$ on a dense set is the same as saying $(f(x),g(x)) \subseteq \Delta$, where $\Delta$ is the diagonal for $x \in U$. But $U$ is dense in $X$, so the diagonal mapping $X \mapsto \Delta \subset X \times X, x \mapsto (x,x)$ is closed.
3. What is the condition on b)? Try with $X$ not irreducible (or even disjoint).